IIT JEE , AIEEE Forum - Mathematics , Algebra

dixit_se

if f(x) is a quadratic expression such that f(x)>0 for all x belongs to R and if g(x)=f(x)+f '(x)+f ''(x), then prove that g(x)>0 for all x belongs to R

gcch29

f(x)>0 & f(x) is quad.

it means D<0 ,a>0

let f(x) = ax^2 +bx +c

f'(x)=2ax+b

f''(x)=2a

D=b^2 - 4ac <0 ---(1)

g(x)=ax^2 +x(b+2a)+(c+b+2a)

it,s discriminant=b^2 -4ac -4a^2

from (1)

D of g(x) is less than 0

since quantity less than zero - (+ive quantity)

thefore g(x) has no real roots

& a>0

thefore it will always lie above x-axis

hence g(x)>0 for all x belonging to real numbers

amar.gupta

Dear,

Let f(x) = ax²+bx+c

f(x) > 0 for x in R , hence a>0 and (b²-4ac)<0

f'(x) = 2ax+b

and f''(x) = 2a

then g(x) = ax²+bx+c + 2ax+b+2a

or g(x) = ax²+(b+2a)x+(c+b+2a)

now g(x) is a quadratic , and since a>0 then we have to prove that :

D = (b+2a)²-4a(c+b+2a)<0

or D = (b²+4a²+4ab)-(4ac+4ab+8a²)

or D =(b²-4ac)-4a²

since (b²-4ac) <0

so D<0

hence g(x) >0 for x in R

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